# Difference between revisions of "2010 AMC 12B Problems/Problem 16"

The following problem is from both the 2010 AMC 12B #16 and 2010 AMC 10B #18, so both problems redirect to this page.

## Problem 16

Positive integers $a$, $b$, and $c$ are randomly and independently selected with replacement from the set $\{1, 2, 3,\dots, 2010\}$. What is the probability that $abc + ab + a$ is divisible by $3$?

$\textbf{(A)}\ \dfrac{1}{3} \qquad \textbf{(B)}\ \dfrac{29}{81} \qquad \textbf{(C)}\ \dfrac{31}{81} \qquad \textbf{(D)}\ \dfrac{11}{27} \qquad \textbf{(E)}\ \dfrac{13}{27}$

## Solution 1

We group this into groups of $3$, because $3|2010$. This means that every residue class mod 3 has an equal probability.

If $3|a$, we are done. There is a probability of $\frac{1}{3}$ that that happens.

Otherwise, we have $3|bc+b+1$, which means that $b(c+1) \equiv 2\pmod{3}$. So either $$b \equiv 1 \pmod{3}, c \equiv 1 \pmod{3}$$ or $$b \equiv 2 \pmod {3}, c \equiv 0 \pmod 3$$ which will lead to the property being true. There are a $\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}$ chance for each bundle of cases to be true. Thus, the total for the cases is $\frac{2}{9}$. But we have to multiply by $\frac{2}{3}$ because this only happens with a $\frac{2}{3}$ chance. So the total is actually $\frac{4}{27}$.

The grand total is $$\frac{1}{3} + \frac{4}{27} = \boxed{\text{(E) }\frac{13}{27}.}$$

## Solution 2 (Minor change from Solution 1)

Just like solution 1, we see that there is a $\frac{1}{3}$ chance of $3|a$ and $\frac{2}{9}$ chance of $3|1+b+bc$

Now, we can just use PIE (Principals of Inclusion and Exclusion) to get our answer to be $\frac{1}{3}+\frac{2}{9}-\frac{1}{3}\cdot\frac{2}{9} = \boxed{E) \frac{13}{27}}$

-Conantwiz2023

## Solution 3 (Fancier version of Solution 1)

As with solution one, we conclude that if $a\equiv0\mod 3$ then the requirements are satisfied. We then have: $$a(bc+c)+a\equiv0 \mod 3$$ $$a(bc+c)\equiv-a \mod 3$$ $$c(b+1)\equiv-1 \mod 3$$ $$b+1\equiv \frac{-1}{c} \mod 3$$ Which is true for one $b$ when $c\not\equiv 0 \mod 3$ because the integers$\mod 3$ form a field under multiplication and addition with absorbing element $0$.

This gives us $P=\frac{1}{3}+\frac{4}{27}=\boxed{\text{(E) }\frac{13}{27}}$.

~Snacc

~IceMatrix

## See also

 2010 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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